[LeetCode]295. Find Median from Data Stream

Description:

Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value.

Examples: 

[2,3,4] , the median is 3

[2,3], the median is (2 + 3) / 2 = 2.5

Design a data structure that supports the following two operations:

• void addNum(int num) - Add a integer number from the data stream to the data structure.
• double findMedian() - Return the median of all elements so far.

For example:

addNum(1)addNum(2)findMedian() -> 1.5addNum(3) findMedian() -> 2

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Solution:

题意：在可变数组下，高效地寻找这一组数的中位数。

思路：

【TLE】①使用插入排序，每次addNum时将新加进来的数插入到已经排好序的vector当中。

时间复杂度：O(logn) + O(n) ~ O(n)。其中O(logn)是用二分查找法寻找插入位置的时间，O(n)是讲n后面的数往后挪一位的时间。

【TLE】②使用快速排序，每次addNum时将新加入的数放到vector的最后，然后用快排队整个vector进行排序。

时间复杂度：O(nlogn)

【AC】③使用双堆排序，priority_queue，分别是大堆排序和小堆排序。

时间复杂度：O(5*logn)。其中5指推入和推出堆的次数，logn指推入和推出堆的时间。结合代码比较容易理解。

class MedianFinder {private:    priority_queue<int, vector<int>, less<int>> lowQueue;    priority_queue<int, vector<int>, greater<int>> highQueue;public:    /** initialize your data structure here. */    MedianFinder() {    }        void addNum(int num) {        lowQueue.push(num);        highQueue.push(lowQueue.top());        lowQueue.pop();                // balanced        if (lowQueue.size() < highQueue.size()) {            lowQueue.push(highQueue.top());            highQueue.pop();        }    }        double findMedian() {        return highQueue.size() == lowQueue.size() ? (double) (highQueue.top() + lowQueue.top()) / 2 : (double) lowQueue.top();    }};/** * Your MedianFinder object will be instantiated and called as such: * MedianFinder obj = new MedianFinder(); * obj.addNum(num); * double param_2 = obj.findMedian(); */